Projectile Questions

1. Sep 8, 2013

KingPapaya

1. The problem statement, all variables and given/known data

We did an experiment where we threw a tennis ball far and high.
For the far toss:
initial height=1.3m
time it took to land=1.92sec
distance traveled=23.4m

For the high toss:
initial height=1.37m
time it took to land=2.61sec
distance traveled=9.5m

Now I have to calculate the maximum height for both tosses.
Then I have to solve for the xmax at maximum height for the long toss.

2. Relevant equations
I know that I have to use delta y=VyoT-.5gt^2 for maximum height.

3. The attempt at a solution
I know the formula to solve for maximum height but there isn't any angle that can be used so I don't know what to do. I tried setting Vyo equals to 0 but that gave me a negative number. For the xmax, I know it is supposed to be less than half of the distance traveled because of the initial height but I don't know how to find it.

2. Sep 8, 2013

voko

If you only care about the max height, ignore the horizontal distance part. Focus solely on the vertical motion. The equation you have is good. What is known and unknown in it?

3. Sep 8, 2013

TSny

Hello, KingPapaya. Welcome to PF!

Consider the far toss. You know the equation that relates y and t. Apply that equation for t = 1.92 s. What can you deduce from that equation?

Repeat for the equation that relates x and t.

[Oops, I see voko already replied while I was constructing my reply.]

4. Sep 8, 2013

KingPapaya

Delta y is y-yo so I have yo, which is 1.3m for far/long and 1.37m for high.
I have t, which is 1 .92 sec for far/long and 2.61 sec for high.
G is 9.8m/s^2.
I'm supposed to be solving for y but I'm not sure what Vyo is

5. Sep 8, 2013

voko

Since the ball hits the ground, do you think you don't have $\Delta y$?

6. Sep 8, 2013

KingPapaya

I thought I have Δy because I have yo

7. Sep 8, 2013

voko

Thus you have everything in that equation - except $v_{0y}$. Can you find $v_{0y}$ then?

8. Sep 8, 2013

KingPapaya

So Vyo=(Δy+.5gt^2)/t?

Vyo=(y-1.3m+.5(9.8ms^2)(1.92sec))/1.92sec

Wait so what do I do about y, since it is also unknown

9. Sep 8, 2013

voko

Where is the ball at t = 1.92 s?

10. Sep 8, 2013

KingPapaya

Oh so since it is on the ground, Δy is -1.3m because 0-1.3m?

11. Sep 8, 2013

voko

Correct!

12. Sep 8, 2013

KingPapaya

After plugging in the numbers to the equation the number I get for long and high are the maximum heights?
Vyo=maximum height?

13. Sep 8, 2013

voko

$v_{0y}$ is the initial vertical velocity.

What is the vertical velocity at the max height?

14. Sep 8, 2013

KingPapaya

Would it be 0?

15. Sep 8, 2013

voko

That is correct as well. Now that you know the initial and the final velocities, and you know the acceleration as well, can you find the time?

16. Sep 8, 2013

KingPapaya

0=Vyo-gt?
So T is (0-Vyo)/-g?

17. Sep 8, 2013

voko

Yes.

18. Sep 8, 2013

KingPapaya

Now what should I do with the time and Vyo solved for to get the maximum height?
Will I have to use the original equation?

19. Sep 8, 2013

voko

Is there any reason you cannot?

20. Sep 8, 2013

KingPapaya

No.
So after I do the same thing for the high toss, should I make a triangle?

Last edited: Sep 8, 2013